Chapter 2 Finite Automata

1
Chapter 2Finite Automata

2
A Finite-State Machine (1)

• A finite state machine is a mathematical model of
  a system, with discrete inputs and outputs

3
A Finite-State Machine (2)

• Finite Automata FA
• a finite set of states
• a set of transitions (edges)
• a start state
• a set of final states
• Defining a FA is a kind of programming.
• Problem definition
• Includes defining possible actions accepting
  condition.
• States ? structure of program
• Includes designating which are initial final.
• Transitions ? program

4
2.1 Deterministic Finite Automata (1)

• DFA M (Q, S, d, q0, F)
• Q a finite set of states
• S a finite set called the alphabet
• d transition function
• total function Q ? S ? Q
• q0 start state q0 ? Q
• F final or accepting states F ? Q

5
Deterministic Finite Automata (2)

• DFA M
• Q q0, q1
• S a, b
• F q1
• The transition function d is given in a tabular
  form called the transition table
• d(q0, a) q1 d(q0, b) q0
• d(q1, a) q1 d(q1, b) q0

6
Deterministic Finite Automata (3)

• A DFA M can be considered to be a language
  acceptor.
• The language of M, L(M), is the set of strings S
  accepted by M.
• A DFA M reads an input string from left to right.
• The next state depends on the current state and
  the unread (unprocessed) symbol.

7
Deterministic Finite Automata (5)

• The DFA M accepts the set of strings over a, b
  that contain the substring bb
• M Q q0, q1, q2, S a, b, F q2
• The transition function d is given in a tabular
  form called the transition table
• d(q0, a) q0 d(q0, b) q1
• d(q1, a) q0 d(q1, b) q2
• d(q2, a) q2 d(q2, b) q2
• Is abba ? L(M)? Yes, since the computation halts
  in state q2, which is a final state
• Is abab ? L(M)? No, since the computation halts
  in state q1, which is NOT a final state

8
Deterministic Finite Automata (6)

• Extended transition function d of a DFA with
  transition function d is a function from Q ? S ?
  Q defined recursively on the length of the input
  string w
• Basis w 0. Then w e and d(qi, e) qi
• Recursive step Let w ? 1. Then d(qi,
  av) d(d(qi, a), v)
• ? qi ? Q, ? a ? S, ? v ? S

9
Deterministic Finite Automata (7)

• A string w is accepted if d(q0, w) ? F
• The language of a DFA M is
• L(M) w ? S d(q0, w) ? F
• DFA M (Q, S, d, q0, F) accepts w ? S ?
• d(q0, w) ? F

10
Deterministic Finite Automata (8)

• Two possibilities for DFA M running on w
• M accepts w
• M accepts w iff the computation of M on w ends up
  (halts) in an accepting configuration
• d(q0, w) ? F
• M rejects w
• M rejects w iff the computation of M on w ends up
  (halts) in a rejecting configuration
• d(q0, w) ? F

11
State Diagrams and Examples (1)

• The state diagram of a DFA M (Q, S, d,q0, F) is
  a labeled graph G defined by the following
  conditions
• The nodes of G are the elements of Q
• The labels on the arcs of G are elements of S
• q0 is the start node, denoted by
• F is the set of accepting nodes, denoted by
• There is an arc from node qi to qj labeled a if
  d(qi, a) qj
• For every node qi and symbol a ? S, there is
  exactly one arc labeled a leaving qi

a
q0
12
State Diagrams and Examples (2)

• Deterministic Finite Automata DFA
• all outgoing edges are labelled with an input
  character
• no state has e- transition, transition on input e
  
• no two edges leaving a given state have the same
  label
• for each state s and input symbol a, there is at
  most one edge label a leaving s
• Therefore the next state can be determined
  uniquely, given the current state and the current
  input character

13
State Diagrams and Examples (1)

• strings over a,b with at least 3 as

14
State Diagrams and Examples (2)

• strings over a,b with length mod 3 0

15
State Diagrams and Examples (3)

• strings over a,b without 3 consecutive as
• A simple example of strings not of the form .

16
State Diagrams and Examples (4)

• Draw a state diagram for DFA M that accepts the
  set of strings over a, b that contain the
  substring bb
• The string ababb is accepted since the halting
  state is the accepting state q2

17
State Diagrams and Examples (5)

• The DFA
• accepts (bab)(ae)
• the set of strings over a, b that do not
  contain the substring aa

b
S
1 a
2 a
0 a
a
a
b
18
State Diagrams and Examples (6)

• The language anbn, n?0 is not regular, so we
  can not build a DFA that accept this language
• It needs an infinite number of states
• But anbn, 1 ? n ? 3 is regular and its DFA is

This DFA is NOT Complete
a
a
a
q0
q1
q2
q3
b
b
b
q4
q5
q6
b
b
19
State Diagrams and Examples (7)

• strings over a, b that contain the substring bb
  OR do not contain the substring aa
• This language is the union of the languages of
  the previous examples

a
1 a
2 a
0 a
a
a
2 a
b
a
b
a
b
b
1 b
2 b
b
S
20
State Diagrams and Examples (8)

• strings over a, b that contain an even number
  of as AND an odd number of bs

b
ea, eb
ea, ob
b
a
a
a
a
oa, eb
oa, ob
b
b
21
State Diagrams and Examples (9)

• Let M be the DFA previous slide
• A DFA M that accepts all strings over a, b
  that do not contain an even number of as AND an
  odd number of bs is shown below
• L(M) a, b - L(M) S - L(M)
• Any string accepted by M is rejected by M and
  vice versa

b
ea, eb
ea, ob
b
a
a
a
a
oa, eb
oa, ob
b
b
22
State Diagrams and Examples (10)

• Let S 0, 1, 2, 3. A string in S is a
  sequence of integers from S
• The DFA M determines whether the sum of elements
  of a string is divisible by 4
• The string 12302 and 0130 should be accepted and
  0111 rejected by M

0
0
w mod 4 0
w mod 4 1
1
3
1
3
3
1
2
2
2
2
w mod 4 2
1
w mod 4 3
0
3
0
23
State Diagrams and Examples (11)

• DFA M1 accepts (ab)c
• M1 is incomplete determinism
• The string abcc is rejected since M1 is unable to
  process the final c from state q2

a
q0
q1
b
c
q2
M1
24
State Diagrams and Examples (12)

• M2 accepts the same language as M1 in previous
  example (ab)c
• The state qe is the error state (dead end)

a
q0
q1
b
c
a, c
b
q2
qe
a, b, c
M2
S
25
State Diagrams and Examples (13)

• strings over a,b with next-to-last symbol a

26
State Diagrams and Examples (14)
What Language is Accepted?
27
2.2 Nondeterministic Finite Automata (1)
28
Nondeterministic Finite Automata (2)

• NFA Formal Definition
• NFA M (Q, S, d, q0, F)
• Q a finite set of states
• S a finite set alphabet
• d transition function
• total function Q ? S ? ?(Q) 2Q - power set of
  Q
• q0 initial/starting state q0 ? Q
• F final or accepting states F ? Q

29
Nondeterministic Finite Automata (3)
Q 0, 1, 2, 3 q0 0 F 3 ? a, b
What Language is defined ?
What is the Transition Table ?
i n p u t
a
b
state
0
0, 1
0
1
?
2
2
?
3
30
Nondeterministic Finite Automata (4)

• Change in d
• For an DFA M, d(q, a) results in one and only one
  state for all states q and alphabet a
• For an NFA M, d(q, a) can result in a set of
  states, zero, one, or more states

qi
a
a
qn
qi
qn
qn
qj
a
qk
a
d(qn, a) qi
d(qn, a) qi, qj , qk
d(qn, a) ?
31
Nondeterministic Finite Automata (5)
a,b
a,b
1
2
3
4
5
a
a
a
b

• Why is this only an NFA and not an DFA?

32
Nondeterministic Finite Automata (6)

• Computing with NFAs
• Computations are different
• We always start from start state. Call it the
  root of the computation.
• Then we might go to different states on one
  symbol.
• Then from those states to new sets of states,
  creating a tree-like computation.
• If one path ends up in a final state, then
  ACCEPT, else REJECT

33
Nondeterministic Finite Automata (7)

• Given an input string, we trace moves
• If no more input in final state, ACCEPT

EXAMPLE Input ababb
Path 1 0 -gt 0 -gt 0 -gt 0 -gt 0 -gt 0 (REJECT) Path
2 0 -gt 0 -gt 0 -gt 1 -gt 2 -gt 3 (ACCEPT)
34
Nondeterministic Finite Automata (8)

• Extended transition function d of a NFA with
  transition function d is a function from Q ? S ?
  2Q (power set) defined recursively on the length
  of the input string w
• Basis w 0. Then w e and d(qi, e) qi
• Recursive step Let w ? 1. Then d(qi,
  av) U d(qj, v), qj ? d(qi, a)
• ? qi ? Q, ? qj ? Q ,? a ? S, ? v? S
• The language of a NFA M is
• L(M) w ? S d(q0, w) ? F ? ?

35
NFA (9)

• The state diagram DFA M1 and NFA M2 accepts
  (ab)bb(ab)

a, b
a
36
NFA (10)

• An NFA that accepts string over a, b with
  substring aa or bb
• There are 2 distinct acceptance paths for the
  string abaaabb

a, b
a, b
start
a
a
2
b
a, b
b
4
37
NFA (11)

• The state diagram DFA M1 and NFA M2 accepts
  (ab)abba(ab)

a, b
a
b
start
b
a
M1
b
a
4
a
b
a, b
a, b
start
b
a
M2
b
a
4
38
NFA with e-Transitions (1)

• NFA with e-transitions, denoted by NFA-e
• Formal Definition
• NFA-e M (Q, S, d, q0, F)
• Q a finite set of states
• S a finite set alphabet
• d transition function
• total function Q ? (S ? e) ? ?(Q)
• q0 initial/starting state q0 ? Q
• F final or accepting states F ? Q

39
e-Transitions (2)

• The input string w is accepted using NFA-e if
  there is a computation that processes the entire
  string and halts in an accepting state
• A computation may continue using e-transition
  after the input string has been completely
  processed

40
e-Transitions (3)

• L(M1) (ab)bb(ab), L(M2) (bab)(ae)
• L(M) L(M1) ? L(M2)

b
a, b
a, b
a
q2,0
q2,1
q1,1
q1,2
b
b
q1,0
M2
M1
b
a, b
a, b
q1,1
q1,2
b
b
q1,0
?
M
b
?
a
q2,0
q2,1
b
41
e-Transitions (4)

• L(M1) (ab)bb(ab), L(M2) (bab)(ae)
• L(M) L(M1)L(M2)

b
a, b
a, b
a
q2,0
q2,1
q1,1
q1,2
b
b
q1,0
M2
M1
b
b
a, b
a, b
a
q1,1
q1,2
b
b
q1,0
q2,0
?
q2,1
M
b
42
e-Transitions (5)

• M accepts all strings over a, b of even length

e
q1
q2
a, b
q0
a, b
e
43
e-Transitions (6)

• Let M1 and M2 be 2 NFA-es as follows

q1,0
q1,f
M1
q2,0
q2,f
M2
M1 (Q1, S, d1, q1,0, F1) M2 (Q2, S, d2,
q2,0, F2)
44
e-Transitions (7)

• L(M) L(M1) ? L(M2)

q1,0
q1,f
M1
?
q0
?
q2,0
q2,f
M2
M (Q1 ? Q2 ? q0, S, d, q0, F1 ? F2) d is the
same as in d1 and d2, but add d(q0, ?) q1,0,
q2,0
45
e-Transitions (8)

• L(M) L(M1)L(M2)

?
q1,0
q2,0
q1,f
M1
q2,f
M2
M (Q1 ? Q2, S, d, q1,0, F2) d is the same as in
d1 and d2, but add d(q, ?) q2,0, where q ?
F1
46
e-Transitions (9)

• L(M) L(M1)

M (Q1 ? q0, qf, S, d, q0, qf) d is the same
as in d1, but add d(q0, ?) q1,0 ,qf d(q1,f,
?) q1,0 ,qf, where q1,f? F1
47

e-Transitions (10)
Given the regular expression a(bc) a( b
c e) Find a transition diagram NFA-? that
recognizes it.
48
e-Transitions (11)
a (bc)
a( b c e)
Now that you have the individual diagrams, or
them as follows
49
e-Transitions (12)
50
e-Transitions (12)

• See examples 2.7, 2.8, 2.9, and 2.10

51
2.3 Equivalence of DFA NFA (1)

• Is NFA more powerful than DFA? NO!
• NFA inefficient to implement directly, so convert
  to a DFA that recognizes the same strings
• Is there a language accepted by an NFA that is
  not accepted by any DFA? No
• NFA is equivalent to DFA.
• Each state in DFA corresponds to a SET of states
  of the NFA

52
Removing Nondeterminism (2)

• ?-closure
• In an NFA-?, the ?-closure(q) of a state q is the
  set of all states that can be reached from q by
  following a path whose edges are all labeled by ?.

a
Start
?
?
q0
q1
q3
?-closure(q0) q0, q1, q2, q3 ?-closure(q1)
q1, q3 ?-closure(q2) q1, q2, q3
?
?
b
b
q2
53
Removing Nondeterminism (3)
NFA-e N (Q, ?, d, q0, F)
?-closure(q) q ? Q
set of NFA-e states that are reachable from q
on ?-transitions only ?-closure(T) T ? Q
NFA-e states reachable from
some state t ? T on
?-transitions only move(T, a) T ? Q, a ?
? set of states to
which there is a transition on
input a from some state t ? T move(q1, q2, ,
qi, a) ?(q1, a) ? ?(q2, a) ? ? ?(qi, a)
No input is consumed
These 3 operations are utilized by algorithms /
techniques to facilitate the conversion process
54
Removing Nondeterminism (4)

• ?-closure(0) 0, 1, 2, 4, 7 (all states
  reachable from 0 on ?-moves)
• ?-closure(0, 6) ?-closure(0) ? ?-closure(6)
• 0, 1, 2, 4, 7 ? 6, 1, 2, 4,7 0, 1,
  2, 4, 6, 7
• move(0, 1, 2, 4, 7, a) 3, 8 (since move(2,
  a)3 and move(7, a)8)

55
NFA-e ?DFA with Subset Construction
NFA-e N ? DFA M construction Given N (Q, S, d,
q0, F), define M (Q, S, d, q0, F) Q ?
P(Q), all the possible NFA-e states M could be
in. q0 e-closure(q0) F q ? Q q ? F
? ? d(q, a) e-closure(move(q, a))

• DFA M has a state ERR ? Q P(Q)
• serves as the error state, when needed
• ?(ERR, a) ERR, ?a ? ? By definition of ?

56
Subset Construction Algorithm Concepts
Start with NFA-e
(a b)abb
First we calculate ?-closure(0) (i.e., state
0) ?-closure(0) 0, 1, 2, 4, 7 (all states
reachable from 0 on ?-moves) Let A 0, 1, 2,
4, 7 be a state of new DFA M
57
Conversion Example (1)
2nd , we calculate a ?-closure(move(A, a))
and b
?-closure(move(A, b)) a ?-closure(move(A, a))
?-closure(move(0,1,2,4,7, a)) adds 3, 8
(since move(2, a) 3 and move(7, a) 8) From
this we have ?-closure(3, 8)
1,2,3,4,6,7,8 (since 3?6?1?4, 6?7, and 1?2
all by ?-moves) Let B 1,2,3,4,6,7,8 be a new
state in D. Define ?M(A, a) B
b ?-closure(move(A,b)) ?-closure(move(0,1,2,4
,7, b)) adds 5 (since move(4, b) 5)
From this we have ?-closure(5)
1,2,4,5,6,7 (since 5?6?1?4, 6?7, and 1?2 all
by ?-moves) Let C 1,2,4,5,6,7 be a new state
in D. Define ?M(A, b) C
58
Conversion Example (2)
3rd, we calculate for state B on a, b a
?-closure(move(B, a)) ?-closure(move(1,2,3,4,6,
7,8, a)) 1,2,3,4,6,7,8 B Define ?M(B, a)
B b ?-closure(move(B, b))
?-closure(move(1,2,3,4,6,7,8, b))
1,2,4,5,6,7,9 D Define ?M(B, b) D
4th, we calculate for state C on a, b a
?-closure(move(C, a)) ?-closure(move(1,2,4,5,6,
7, a)) 1,2,3,4,6,7,8 B Define ?M(C, a)
B b ?-closure(move(C, b)) ?-closure(move(1,2
,4,5,6,7, b)) 1,2,4,5,6,7 C Define ?M(C,
b) C
59
Conversion Example (3)
5th , we calculate for state D on a, b a
?-closure(move(D, a)) ?-closure(move(1,2,4,5,6,
7,9, a)) 1,2,3,4,6,7,8 B Define ?M(D, a)
B b ?-closure(move(D, b))
?-closure(move(1,2,4,5,6,7,9, b))
1,2,4,5,6,7,10 E Define ?M(D, b) E
Finally, we calculate for state E on a, b a
?-closure(move(E, a)) ?-closure(move(1,2,4,5,6,
7,10, a)) 1,2,3,4,6,7,8 B Define ?M(E,
a) B b ?-closure(move(E, b))
?-closure(move(1,2,4,5,6,7,10, b))
1,2,4,5,6,7 C Define ?M(E, b) C
60
Conversion Example (4)
This gives the transition table for the DFA M of
b
b
b
a
b
b
a
start
a
a
a
61
Converting NFA-e to DFA 2nd Example
RE c(ab)c
1st we calculate ?-closure(0) ?-closure(0)
0, 1, 2, 6, 8 (all states reachable from 0 on
?-moves) Let A0, 1, 2, 6, 8 be a state of new
DFA M
62
Conversion Example (1)
2nd , we calculate a ?-closure(move(A, a))
and b
?-closure(move(A, b)) c
?-closure(move(A, c)) a ?-closure(move(A, a))
?-closure(move(0,1,2,6,8, a)) adds 3
(since move(2, a) 3) From this we have
?-closure(3) 3 (since 3?3 by ?-moves) Let
B 3 be a new state. Define ?M(A, a) B
b ?-closure(move(A,b)) ?-closure(move(0,1,2,6
,8, b)) There is NO transition on b for all
states 0,1,2,6 and 8 Define ?M(A, b) Reject
c ?-closure(move(A, c)) ?-closure(move(0,1,2,
6,8, c)) adds 7 (since move(6, c) 7) From
this we have ?-closure(7)
1,2,5,6,7,8 (since 7?5?8, 7?5?1?2, and 7?5?1?6
by ?-moves) Let C 1,2,5,6,7,8 be a new state.
Define ?M(A, c) C
63
Conversion Example (2)
3rd , we calculate a ?-closure(move(B, a))
and b
?-closure(move(B, b)) c
?-closure(move(B, c)) a ?-closure(move(B, a))
?-cclosure(move(3, a)) There is NO
transition on a for state 3 Define ?M(B, a)
Reject
b ?-closure(move(B, b)) ?-closure(move(3,
b)) adds 4 (since move(3, b) 4) From this we
have ?-closure(4) 1,2,4,5,6,8 (since
4?5?8, 4?5?1?2, and 4?5?1?6 by ?-moves) Let D
1,2,4,5,6,8 be a new state. Define ?M(B, b)
D
c ?-closure(move(B, c)) ?-closure(move(3,
c)) There is NO transition on c for state 3
Define ?M(B, c) Reject
64
Conversion Example (3)
4th , we calculate a ?-closure(move(C, a))
and b
?-closure(move(C, b)) c
?-closure(move(C, c)) a ?-cclosure(move(C,
a)) ?-closure(move(1,2,5,6,7,8,a)) adds 3
(since move(2, a) 3) ?-closure(3)
3 Define ?M(C, a) B
b ?-closure(move(C, b)) ?-closure(move(1,2,5,
6,7,8,b)) There is NO transition on b for all
states 1,2,5,6, and 7 Define ?M(C, b) Reject
c ?-closure(move(C, c)) ?-closure(move(1,2,5,
6,7,8,c)) adds 7 (since move(6, c) 7) From
this we have ?-closure(7)
1,2,5,6,7,8 (since 7?5?8, 7?5?1?2, and 7?5?1?6
by ?-moves) Define ?M(C, c) C
65
Conversion Example (4)
5th , we calculate a ?-closure(move(D, a))
and b
?-closure(move(D, b)) c
?-closure(move(D, c)) a ?-closure(move(D, a))
?-closure(move(1,2,4,5,6,8,a)) adds 3
(since move(2, a) 3) ?-closure(3)
3 Define ?M(D, a) B
b ?-closure(move(D, b)) ?-closure(move(1,2,4,
5,6,8, b)) There is NO transition on b for all
states 1,2,4,5,6, and 8 Define ?M(D, b) Reject
c ?-closure(move(D, c)) ?-closure(move(1,2,4,
5,6,8,c)) adds 7 (since move(6, c) 7) From
this we have ?-closure(7)
1,2,5,6,7,8 (since 7?5?8, 7?5?1?2, and 7?5?1?6
by ?-moves) Define ?M(D, c) C
66
The Resulting DFA M
Which States are FINAL States ?
67
NFA-e ? DFA Example

• strings over a,b with next-to-last symbol a

a/b
d a b e
q0 q0, q1 q0 ?
q1 q2 q2 ?
q2 ? ? ?
a
a/b
q2
q1
q0
M (q0, q1, q2, a,b, d, q0, q2)
b
q0 q1
q0
b
b
q0 q2
q0 q1 q2
b
Equivalent DFA
68
An Example of NFA-e ? DFA

• Consider a simple NFA
• Construct a corresponding DFA

69
Examples

• See examples 2.11, 2.12, and 2.13

70
Class Discussion

• Construct a DFA equivalent to this NFA

0
71
Class Discussion
Notice that the state with label 0, 1, 2 is
from the set of states given by the
nondeterministic transition ?(0, a) 0, 1, 2.
Also notice that any state whose label contains
an accepting state is defined as an accepting
state in the deterministic machine.
72
2.4 Reduction of the Number of States in FA

• Self study
• but not included in the exams material